Answer:
The second dart leaves the gun two times as faster than the first one.
Explanation:
Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle
![U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}.](https://tex.z-dn.net/?f=%20U_s%20%3D%20K_d%20%5C%5C%20%5Cfrac%7B1%7D%7B2%7D%20kx%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%5C%5C%20v%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7Dx%5E2%7D.%20)
Given an arbitrary
and its double,
, launch velocities are
![v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}.](https://tex.z-dn.net/?f=%20v_1%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7Dx%5E2%7D%20%5Ctext%7B%20and%7D%20%5C%5C%20v_2%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%5Cleft%282x%5Cright%29%5E2%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D4x%5E2%7D%20%3D%202%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7Dx%5E2%7D%20%3D%20%5Cmathbf%7B2v_1%7D.%20)
Answer:
2.8 m 7.4 m/s
Explanation:
write all the values then use the equations of motion to find the distance and speed. please see attached photo
Answer:
85.556metres
Explanation:
Using pythagorean theorem
C²=A²+B²
we have c as the hypotenuse vector A thus:
93.8²=A²+38.4²
93.8²-38.4²=A²
8794.44-1474.56=A²
7319.88=A²
A=85.556
I think the correct answer from the choices listed above is option A. A high frequency wave is a wave with a low level of energy and a high pitch. Frequency is the number of waves passing per second of time. Hope this answers the question.
I think it's A) Sunspots. I hope this helps:)